The equation of a circle is \( (x+2)^{2}+(y-3)^{2}=9 \). What are the coordinates of the center of this circle?

To find the coordinates of the center of the circle given by the equation \[ (x+2)^{2}+(y-3)^{2}=9, \] we can identify the center from the standard form of the equation of a circle, which is \[ (x-h)^{2}+(y-k)^{2}=r^{2}, \] where \((h, k)\) is the center of the circle and \(r\) is the radius. From the given equation: - The term \((x+2)^{2}\) indicates that \(h = -2\) (since it is in the form \((x - h)\)). - The term \((y-3)^{2}\) indicates that \(k = 3\). Thus, the coordinates of the center of the circle are \[ (-2, 3). \] Therefore, the center of the circle is at the point \((-2, 3)\).